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In a certain residential area 50% of the residents subscribe to the morning newspapers, 65\% subscribe to afternoon newspapers, and 85% subscribe to at least one of the two newspapers. What percentage of the residents subscribe to both newspapers..
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pqList.priorityInsert(50);
pqList.priorityInsert(40);
pqList.priorityInsert(30);
pqList.priorityInsert(20);
pqList.priorityInsert(10);
cout<<"-------------- PriorityQueue -------------"<<endl;
pqList.displayForward();
cout<<"calling priorityRemove()"<<endl;
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pqList.displayForward();
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}
Sample output"
-------------- PriorityQueue -------------
List (first-->last): 10 20 30 40 50
calling priorityRemove()
Removed node with key 50
List (first-->last): 10 20 30 40
Removed node with key 40
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I may change numbers when grading.
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//doublyLinked.cpp
//demonstrates doubly-linked list
#include <iostream>
using namespace std;
////////////////////////////////////////////////////////////////
class Link {
public:
double dData; //data item
Link * pNext; //next link in list
Link * pPrevious; //previous link in list
public:
//-------------------------------------------------------------
Link(double dd): //constructor
dData(dd), pNext(NULL), pPrevious(NULL) {}
//-------------------------------------------------------------
void displayLink() //display this link
{
cout << dData << " ";
}
//-------------------------------------------------------------
}; //end class Link
////////////////////////////////////////////////////////////////
class DoublyLinkedList {
private:
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Link * pLast; //pointer to last item
public:
//-------------------------------------------------------------
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//-------------------------------------------------------------
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{
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{
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pCurrent = pCurrent -> pNext; //move to next link
delete pOldCur; //delete old current
}
}
//-------------------------------------------------------------
bool isEmpty() //true if no links
{
return pFirst == NULL;
}
//-------------------------------------------------------------
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{
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Implement a priority queue using a doublyLinked.cpp where the node with the highest
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If displayForward() displays List (first-->last) : 10 30 40 55
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Demonstrate by inserting keys at random, displayForward(), call remove then
displayForward() again.
int main()
{
DoublyLinkedList pqList;
pqList.priorityInsert(50);
pqList.priorityInsert(40);
pqList.priorityInsert(30);
pqList.priorityInsert(20);
pqList.priorityInsert(10);
cout<<"-------------- PriorityQueue -------------"<<endl;
pqList.displayForward();
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double key = pqList.priorityRemove();
cout<<"Removed node with key "<<key<<endl;
pqList.displayForward();
key = pqList.priorityRemove();
cout<<"Removed node with key "<<key<<endl;
pqList.displayForward();
return 0;
}
Sample output"
-------------- PriorityQueue -------------
List (first-->last): 10 20 30 40 50
calling priorityRemove()
Removed node with key 50
List (first-->last): 10 20 30 40
Removed node with key 40
List (first-->last): 10 20 30
I may change numbers when grading.
doublyLinked.cpp
//doublyLinked.cpp
//demonstrates doubly-linked list
#include <iostream>
using namespace std;
////////////////////////////////////////////////////////////////
class Link {
public:
double dData; //data item
Link * pNext; //next link in list
Link * pPrevious; //previous link in list
public:
//-------------------------------------------------------------
Link(double dd): //constructor
dData(dd), pNext(NULL), pPrevious(NULL) {}
//-------------------------------------------------------------
void displayLink() //display this link
{
cout << dData << " ";
}
//-------------------------------------------------------------
}; //end class Link
////////////////////////////////////////////////////////////////
class DoublyLinkedList {
private:
Link * pFirst; //pointer to first item
Link * pLast; //pointer to last item
public:
//-------------------------------------------------------------
DoublyLinkedList(): //constructor
pFirst(NULL), pLast(NULL) {}
//-------------------------------------------------------------
~DoublyLinkedList() //destructor (deletes links)
{
Link * pCurrent = pFirst; //start at beginning of list
while (pCurrent != NULL) //until end of list,
{
Link * pOldCur = pCurrent; //save current link
pCurrent = pCurrent -> pNext; //move to next link
delete pOldCur; //delete old current
}
}
//-------------------------------------------------------------
bool isEmpty() //true if no links
{
return pFirst == NULL;
}
//-------------------------------------------------------------
void insertFirst(double dd) //insert at front of list
{
Link * pNewLink = new Link(dd); //make new link
if (isEmpty()) //if empty list,
pLast = pNewLink; //newLink <-- last
else
pFirst -> pPrevious = pNewLink; //newLi.
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